Chapter 15 - Differentiation in Several Variables - 15.8 Lagrange Multipliers: Optimizing with a Constraint - Exercises - Page 832: 34

We show that for both of the following two problems: Problem 1. Minimize surface area $S$ for fixed volume $V$ Problem 2. Maximize volume $V$ for fixed surface area $S$ $P = \left( {r,h} \right)$ is a Lagrange critical point if $h=2r$. Using the contour plots in Figure 19 we explain why - $S$ has a minimum for fixed $V$ but no maximum (corresponding to Problem 1) and - $V$ has a maximum for fixed $S$ but no minimum (corresponding to Problem 2)

Problem 1. Minimize surface area $S$ for fixed volume $V$ We have the surface area (including the top and bottom): $S\left( {r,h} \right) = 2\pi rh + 2\pi {r^2}$ Our task is to minimize $S$ subject to a fixed volume constraint $V\left( {r,h} \right) = \pi {r^2}h = {V_0}$. Step 1. Write out the Lagrange equations Using Theorem 1, the Lagrange condition $\nabla S = \lambda \nabla V$ yields $\left( {2\pi h + 4\pi r,2\pi r} \right) = \lambda \left( {2\pi rh,\pi {r^2}} \right)$ So, the Lagrange equations are $2\pi h + 4\pi r = \lambda \left( {2\pi rh} \right)$, ${\ \ \ }$ $2\pi r = \lambda \left( {\pi {r^2}} \right)$ (1) ${\ \ \ \ }$ $h + 2r = rh\lambda $, ${\ \ \ }$ $2 = r\lambda $ Step 2. Solve for $\lambda$ in terms of $r$ and $h$ The second equation of (1) implies that $r \ne 0$ and $\lambda \ne 0$. As a result $h \ne 0$. From equation (1) we obtain $\lambda = \frac{2}{r} = \frac{{h + 2r}}{{rh}}$ $2 = 1 + 2\frac{r}{h}$ $1 = 2\frac{r}{h}$ So, $h = 2r$. Step 3. Solve for $r$ and $h$ using the constraint Substituting $h = 2r$ in the constraint gives $\pi {r^2}\left( {2r} \right) = {V_0}$ ${r^3} = \frac{{{V_0}}}{{2\pi }}$ $r = {\left( {\frac{{{V_0}}}{{2\pi }}} \right)^{1/3}}$ Thus, if $h = 2r$, the Lagrange critical point is $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}},2{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}}} \right)$. Step 4. Calculate the critical values Using the critical point $\left( {r,h} \right) = \left( {{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}},2{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}}} \right)$ we evaluate the extreme value of $S$ subject to the constraint ${V_0} = \pi {r^2}h$: $S\left( {{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}},2{{\left( {\frac{{{V_0}}}{{2\pi }}} \right)}^{1/3}}} \right) = 3{\left( {2\pi {V_0}^2} \right)^{1/3}}$ For a fixed constraint ${V_0}$ we may increase $r$ and at the same time decrease $h$ such that ${V_0}$ is constant. Since $S$ is increasing, we conclude that $3{\left( {2\pi {V_0}^2} \right)^{1/3}}$ is a minimum value of $S$ subject to the constraint. Using the contour plots in Figure 19 (or the figure attached), we see that there is one level curve which touches the constraint curve at the critical point (red point). Those at the left of this level curve do not satisfy the constraint. However, those at the right increase without upper bound. Therefore, we conclude that $3{\left( {2\pi {V_0}^2} \right)^{1/3}}$ is a minimum value of $S$ subject to the fixed volume ${V_0}$, but $S$ has no maximum. Problem 2. Maximize volume $V$ for fixed surface area $S$ We have the volume $V\left( {r,h} \right) = \pi {r^2}h$. Our task is to maximize $V$ subject to a fixed surface area constraint $S\left( {r,h} \right) = 2\pi rh + 2\pi {r^2} = {S_0}$. Step 1. Write out the Lagrange equations Using Theorem 1, the Lagrange condition $\nabla V = \lambda \nabla S$ yields $\left( {2\pi rh,\pi {r^2}} \right) = \lambda \left( {2\pi h + 4\pi r,2\pi r} \right)$ So, the Lagrange equations are $2\pi rh = \lambda \left( {2\pi h + 4\pi r} \right)$, ${\ \ \ }$ $\pi {r^2} = \lambda \left( {2\pi r} \right)$ (1) ${\ \ \ \ }$ $rh = \lambda \left( {h + 2r} \right)$, ${\ \ \ }$ $r = 2\lambda $ Step 2. Solve for $\lambda$ in terms of $r$ and $h$ Since ${S_0} \ne 0$, we may assume that $r \ne 0$ and $h \ne 0$. From equation (1) we obtain $\lambda = \frac{{rh}}{{h + 2r}} = \frac{r}{2}$ $\frac{h}{{h + 2r}} = \frac{1}{2}$ $2h = h + 2r$ So, $h = 2r$. Step 3. Solve for $r$ and $h$ using the constraint Substituting $h = 2r$ in the constraint gives $2\pi r\left( {2r} \right) + 2\pi {r^2} = {S_0}$ $6\pi {r^2} = {S_0}$ $r = {\left( {\frac{{{S_0}}}{{6\pi }}} \right)^{1/2}}$ Thus, if $h = 2r$, the Lagrange critical point is $\left( {r,h} \right) = \left( {{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}},2{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}}} \right)$. Step 4. Calculate the critical values Using the critical point $\left( {r,h} \right) = \left( {{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}},2{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}}} \right)$ we evaluate the extreme value of $V$ subject to the constraint ${S_0} = 2\pi rh + 2\pi {r^2}$: $V\left( {{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}},2{{\left( {\frac{{{S_0}}}{{6\pi }}} \right)}^{1/2}}} \right) = \frac{{{S_0}^{3/2}}}{{3\sqrt {6\pi } }}$ Since $h>0$, for a fixed ${S_0}$ we cannot increase $r$ without violating the constraint. Hence, we conclude that $\frac{{{S_0}^{3/2}}}{{3\sqrt {6\pi } }}$ is a maximum value of $V$ subject to the constraint. Using the contour plots in Figure 19 (or the attached figure), we see that there is one level curve which touches the constraint curve at the critical point (red point). Those at the right of this level curve do not satisfy the constraint. However, those at the left decrease without lower bound. Therefore, we conclude that $\frac{{{S_0}^{3/2}}}{{3\sqrt {6\pi } }}$ is a maximum value of $V$ subject to the fixed surface area ${S_0}$, but $V$ has no minimum.